Question

If the ratio of the sum of m term and n term of an A. P. Is m²:n²then
prove that d=2a​

Answer 1

Step-by-step explanation:

In APs, nth term = a + (n - 1)d, where a is first term and d is common difference.

Let first term of this AP be a and common difference be d.

Sum of n terms is given be (n/2) { 2a + (n - 1)d }

So,

$=>\frac{m^2}{n^2}=\frac{\frac{m}{2}\{2a+(m-1)d\} }{\frac{n}{2}\{2a+(n-1)d\} }$

$=>\frac{m}{n} =\frac{2a+(m-1)d}{2a+(n-1)d}$

=> m[ 2a + (n - 1)d ] = n[ 2a + (m - 1)d ]

=> 2am + mnd - dm = 2an + mnd - nd

= > 2am - dm = 2an - nd

= > 2am - 2an = dm - nd

= > 2a(m - n) = d(m - n)

= > 2a = d

Proved.